NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry are provided here in a detailed and easy way. In this chapter, students are introduced with various important topics and students learn how to calculate the distance between any two points if their coordinates are given, finding the area of a triangle whose three points are given, etc. NCERT class 10 maths solutions for chapter 7 coordinate geometry given here are very simple and are extremely easy to understand and any students can clear their doubts instantly. These given solutions for coordinate geometry will help the students to clear their doubts and have a proper understanding of the concepts taught in this chapter. Exercise-wise PDFs is also available here which the students can download and access all the solutions in offline mode. Check the coordinateÂ geometry NCERT solutions from below.

### NCERT Solutions for Class 10 Maths Chapter 7 Exercises

**1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.**

|AB| = \(\sqrt{(x-5)^{2}+(y-1)^{2}}\)

= \(\sqrt{x^{2}+16-8x+y^{2}+4-4y}\)

|AC| = \(\sqrt{(x+2)^{2}+(y-5)^{2}}\)

= \(\sqrt{x^{2}+4+4x+y^{2}+16-8y}\)

Since, |AB| = |AC|

\(\Rightarrow\) \(x^{2}+y^{2}-8x-4y+20\) = \(x^{2}+y^{2}+4x-8y+20\)

\(\Rightarrow\)Â Â Â Â Â 8y â€“ 4y = 4x + 8x

\(\Rightarrow\)Â Â Â Â Â y = 3x

**Â 2.Â ****Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)**

Let the vertices of the \(\bigtriangleup\) be P(0, 8), Q(0, 0) and R(6, 0)

âˆ´ PQ = \(\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8\)

âˆ´ QR = \(\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6\)

âˆ´ RP = \(\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10\)

âˆ´ Â Perimeter of \(\bigtriangleup\) = 8 + 6 + 10 = 24 units

**3.****Find the coordinates of the perpendicular bisector of the line segment ****joining the points P (1, 4) and Q (2, 3), cuts the y-axis.**

Here, O (0, y), P (1, 4) and Q (2, 3)

AO = BO

\(\Rightarrow\) \(\sqrt{(0-1)^{2}+(y-4)^{2}}\) = \(\sqrt{(0-2)^{2}+(y-3)^{2}}\)

\(\Rightarrow\) \(1+y^{2}+16-8y= 4+y^{2}+9-6y\)

\(\Rightarrow\)Â Â 8y â€“ 6y = 17 -13

\(\Rightarrow\)Â Â 2y = 4

Â Â Â Â Â Â Â Â Â Â Â Â y = 2

âˆ´The required point is (0, 2)

**4. The points which divides the line segment joining the points (5, 8) and (1, 2) in ratio 1:2 internally lies in which quadrant?**

Now C\(\left ( \frac{1+10}{3}, \frac{2+16}{3} \right )\)

âˆ´Â C\(\left ( \frac{11}{3},6 \right )\)

Since, C\(\left ( \frac{11}{3},6 \right )\) lies in IV quadrant.

**5. If A \(\left ( \frac{x}{2},3 \right )\) is the mid-point of the line segment joining the points B(4,1) and C(8,7). Find the value of x.**

âˆ´\(\frac{4+8}{2}=\frac{a}{2}\)

\(\Rightarrow\) 24 = 2a

\(\Rightarrow\) a = 12

**Â **

**6. A line intersects the y-axis at point A and B respectively. If (3, 4) is the mid-point of AB. Find the coordinates of A and B.**

Here, \(\frac{x+0}{2}=3\) \(\Rightarrow x= 6\)

\(\frac{0+y}{2}=4\) \(\Rightarrow x= 8\)

âˆ´ The coordinates of A and B are (0, 8) and (6, 0).

**7. Find the fourth vertex S of a parallelogram PQRS whose three vertices are P (-3, 8), Q (5, 4)Â ****and R (6, 2).**

Here, \(\frac{-3+6}{2}=\frac{5+x}{2}\Rightarrow x=-1\)

\(\frac{8+2}{2}=\frac{4+x}{2}\Rightarrow x=3\)

âˆ´The fourth vertices S of a parallelogram PQRS is S(-1, 3)

**Â 8.Â ****Find the coordinates of the point which is equidistant from the three vertices of the**

\(\bigtriangleup\)POQ.

Point equidistant from the three vertices ofÂ a right angle triangle is the mid-point of hypotenuse.

\(\Rightarrow \left ( \frac{4x+0}{2},\frac{0+4y}{2} \right )\Rightarrow (2x,2y)\)

**Â **

**9. Find the area of a triangle with vertices (x, y + z), (y, z + x) and (z, x + y).**

Area of the required

\(\bigtriangleup = \frac{1}{2}|x(z+x-x-y)+y(x+y-y-z) +(y+z-z-x)|\)

= \(\frac{1}{2}|x(z-y)+y(x-z)+z(y-a)|\)

= \(\frac{1}{2}|0|\) = 0.

**10. Find the area of a triangle with vertices P (2, 0), Q (6, 0) and R (5, 4)**

Area of the required \(\bigtriangleup PQR\)

= \(\frac{1}{2}|2(0 – 2)+ 6(4 – 0)+ 5(0 – 0)|\)

= \(\frac{1}{2}|24-4|=10\) Sq. units.

**11. If the points O (0, 0), P (3, 4), Q (x, y) are collinear, then write the relation between x and y.**

Since,Â O (0, 0), P (3, 4), Q (x, y) are collinear.

\(\Rightarrow\) Area of a triangle formed by these points vanishes

âˆ´Â \( \frac{1}{2}|0(4 – y)+3(y-0)+x(0-4)|=0\)

\(\Rightarrow y-2x=0\Rightarrow 2x=y\)

**Â **

**12. Find the coordinates of the point O dividing the line segment joining the point P (6, 5)**

**and Q(2, 10) in the ratio 2:1.**

\(x=\frac{2(2)+1(6)}{5}= \frac{4+6}{5}= 2\)

and \(y=\frac{2(10)+1(5)}{5}= \frac{20+5}{5}= 5\)

âˆ´ O(2, 5)

**13. The line joining A(4, 3) and B(-3, 6) meets y-axis at C. At what ratio dose C divides the line segment AB?**

Since R lies on the y-axis, let C be (0, y)

Suppose that the required ratio be z:1

âˆ´\(x=\frac{-3z+4}{z+1}\Rightarrow 0= -3z+4\Rightarrow z=\frac{4}{3}\)

âˆ´ The required ratio is 4 : 3.

In coordinate geometry, students are introduced to several important concepts. These concepts are extremely important for CBSE class 10 board exam as almost 6 marks are included in the exam from this chapter. In the coordinate geometry chapter, students are introduced to a lot of important topics and formulas like distance formula, section formula, and area of a triangle. All these concepts are extremely crucial as questions are asked in the class 10 exam. Apart from that, these concepts are also included in some of the important maths topics in the later grades. Students are required to be thorough with all the topics that are included in this chapter and should understand the solved examples properly.

Students also need to solve exercise problems in this chapter and can refer to this NCERT Solutions for Class 10 Maths Coordinate Geometry in case of any doubts. Apart from this, students can also find the solutions for all the chapter by visitingÂ NCERT Solutions For Class 10. Keep visiting BYJUâ€™S to get complete assistance for CBSE class 10 board exams. Students are provided with sample papers, question papers, and notes to help them to prepare for their exams in a more effective way. Students are also suggested to download BYJUâ€™S- The Learning App to get a personalized and engaging learning experience.